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Question Number 14905 by ajfour last updated on 05/Jun/17

Commented by ajfour last updated on 05/Jun/17

$Q . 14797 (c o n s t r u c t i o n m e t h o d)$ $g i v e n D E = a, E B = b; f i n d$ $A r e a o f △ A E C .$ $l e t d i a g o n a l o f s q u a r e = 2 s$ $2 s = a \cos α + b \cos (θ - α)$ $s = \frac{1}{2} (a \cos α + b \cos (θ - α)) . . . (i)$ $l e t ⊥ d i s t a n c e o f E f r o m A C b e h .$ $h = s - b \cos (θ - α)$ $h = \frac{1}{2} (a \cos α - b \cos (θ - α)) . . . (i i)$ $\tan α = \frac{\sin α}{\cos α} = \frac{b \sin θ}{a + b \cos θ}$ $\Rightarrow a \sin α = b (\sin θ \cos α - \cos θ \sin α)$ $o r a \sin α = b \sin (θ - α) . . . (i i i)$ $A r e a o f △ A E C = \frac{1}{2} (2 s) h = s h$ $= \frac{1}{4} [a^{2} \cos^{2} α - b^{2} \cos^{2} (θ - α)]$ $[u s i n g (i) a n d (i i)]$ $= \frac{1}{4} [a^{2} - a^{2} \sin^{2} α - b^{2} + b^{2} \sin^{2} (θ - α)]$ ${A r e a}_{△ A E C} = \frac{1}{4} (a^{2} - b^{2}) .$ $[u s i n g (i i i)] .$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$g o d b l e s s y o u m r A j f o u r . i t i s a b e a u t i f u l$ $p r o v e . i l o v e i t v e r y m u c h . t h a n k s .$
Commented by ajfour last updated on 05/Jun/17

$t h a n k s s i r, i w a s c u r i o u s a b o u t$ $t h e i n d e p e n d e n c e o f t h e c o n c e r n e d$ $a r e a w i t h θ . I t i s s t i l l n o t c l e a r t o$ $m e . .!$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

$f o r w h a t ? t h e a r e a c o m p u t e d f i n a l l y$ $w h i t n o i n d e p e n d e n c e t o θ .$
Commented by ajfour last updated on 05/Jun/17

$i m e a n i i t i s l e n g t h y, n o t d i r e c t$ $o r o b v i o u s a s i t i s t o m r W 1 . .$
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